I have started writing a small piece of code to print all the list of users available in the linux box. But I want to pass one by one user into my command to display each user details together.
to list all users
root#bt# getent passwd | grep /home/ | cut -d ':' -f 1
root
san
postgres
Now I want to pass one by user in to the below command to display each user details together.
root#bt# chage -l ${user1} ; chage -l ${user2} etcc.
should I need to user for loop or while loop here?
can any one help me in suggesting how to write the same?
You can use the while loop:
getent passwd | grep /home/ | cut -d ':' -f 1 | \
while read user ; do
chage -l "$user"
done
or the for loop:
for user in $(getent passwd | grep /home/ | cut -d ':' -f 1) ; do
chage -l "$user"
done
or xargs:
getent passwd | grep /home/ | cut -d ':' -f 1 | \
xargs -n1 chage -l
I would use xargs, which runs a command on each output item of the previous pipe:
getent passwd | grep /home/ | cut -d ':' -f 1 | sudo xargs -I % sh -c '{ echo "User: %"; chage -l %; echo;}'
sudo is used to get information about all users, if you don't have access to this information then you can remove sudo
-I % is used to specify that % is a placeholder for the input item (in your case a user)
sh -c '{ command1; command2; ...;}' is the command executed by xargs on every % item; in turn, the command sh -c allows multiple shell commands to be executed
'{ echo "User: %"; chage -l %; echo;}' echoes the current user in %, then runs chage -l on this user and finished with a final empty echo to format the ouput
Related
how do I set a single password for all users in a Linux system? For example, how will I set a password, x, so that it's the password of all users in the system?
I was thinking of a for loop that iterates between each other but then I realised I have no clue on how to go about this.
You could manually change all user accounts in question with the following, it will prompt you for the new password
$ sudo passwd <username>
You could automate this with a script. Or you could use a convoluted command at the command line, which is what I would do. The below example will pull all users from the passwd file, filter out the users that cannot login, and then run a loop to set their password
using cat piped to grep you can get a list of all users and filter out the users with "nologin" or "false" in their config. If you get users that you do not want, change the filter items or add the username to the grep statement to filter them out, separate each filter item with \|
$ cat /etc/passwd | grep -Ev nologin\|false
using awk you can get just the username to print out
$ cat /etc/passwd | grep -Ev nologin\|false | awk '{split($0,a,":");print a[1]}'
running this command in a for loop will let us run a command on each user, to test just echo the username
$ for user in `cat /etc/passwd | grep -Ev nologin\|false | awk '{split($0,a,":");print a[1]}'`; do echo $user; done
the tricky part is passing a password to the passwd command. switch to the root user and then echo the password to the passwd command. Here is an example
$ sudo -i
# (echo 'newpassword'; echo 'newpassword') | passwd <username>
however you do not want the password in your command line history. put the password in a tempfile and then cat it to xargs. As an example, just echo the password using xargs
$ sudo -i
# vi tempfile
enter only one line with the new password
# cat tempfile | xargs -i echo {}
now you'll use xargs to echo to passwd. this is tricky again because you need to run two echo commands correctly, just tell xargs to run the command in a sub shell
$ sudo -i
# cat tempfile | xargs -i /bin/bash -c "(echo '{}'; echo '{}') | sudo passwd <username>"
now add the xargs command in the for loop
$ sudo -i
# for user in `cat /etc/passwd | grep -Ev nologin\|false | awk '{split($0,a,":");print a[1]}'`; do cat tempfile | xargs -i /bin/bash -c "(echo '{}'; echo '{}') | sudo passwd $user"; done
That should be it, let me know if you have questions
I tested this example on ubuntu 20.04 using GNU bash 5.0.17
i have created 2 sudo users and the users are showing in /etc/sudoers . is there any way to list out all the sudo users in linux .
i tried some commands from google but not worked for me:-
grep '^sudo:.*$' /etc/group | cut -d: -f4
getent group sudo | cut -d: -f4
the user added in /etc/sudoers file is:
kamlesh ALL=(ALL) NOPASSWD: ALL
for usr in $(cut -f 1 -d : /etc/passwd);
do
sudo -l $usr > /dev/null 2&>1 && echo $usr;
done
Extract all the users from /etc/passwd using cut (set field delimiter to ":" with -d and read the first field -f 1) Process each read value into the variable usr using a loop. For each usr, run sudo -l on that user to see if that user can execute any commands with sudo. Redirect stdin and sterr to /dev/null and if the return code is 0, echo the $usr
I'm trying to write a script that builds a list of nodes then ssh into the first node of that list
and runs a checknodes.sh script which it's self is just a for i loop that calls checknode.sh
The first 2 lines seems to work ok, the list builds successfully, but then I get either get just the echo line of checknodes.sh to print out or an error saying cat: gpcnodes.txt: No such file or directory
MYSCRIPT.sh:
#gets the master node for the job
MASTERNODE=`qstat -t -u \* | grep $1 | awk '{print$8}' | cut -d'#' -f 2 | cut -d'.' -f 1 | sed -e 's/$/.com/' | head -n 1`
#builds list of nodes in job
ssh -qt $MASTERNODE "qstat -t -u \* | grep $1 | awk '{print$8}' | cut -d'#' -f 2 | cut -d'.' -f 1 | sed -e 's/$/.com/' > /users/issues/slow_job_starts/gpcnodes.txt"
ssh -qt $MASTERNODE cd /users/issues/slow_job_starts/
ssh -qt $MASTERNODE /users/issues/slow_job_starts/checknodes.sh
checknodes.sh
for i in `cat gpcnodes.txt `
do
echo "### $i ###"
ssh -qt $i /users/issues/slow_job_starts/checknode.sh
done
checknode.sh
str=`hostname`
cd /tmp
time perf record qhost >/dev/null 2>&1 | sed -e 's/^/${str}/'
perf report --pretty=raw | grep % | head -20 | grep -c kernel.kallsyms | sed -e "s/^/`hostname`:/"
When ssh -qt $MASTERNODE cd /users/issues/slow_job_starts/ is finished, the changed directory is lost.
With the backquotes replaced by $(..) (not an error here, but get used to it), the script would be something like
for i in $(cat /users/issues/slow_job_starts/gpcnodes.txt)
do
echo "### $i ###"
ssh -nqt $i /users/issues/slow_job_starts/checknode.sh
done
or better
while read -r i; do
echo "### $i ###"
ssh -nqt $i /users/issues/slow_job_starts/checknode.sh
done < /users/issues/slow_job_starts/gpcnodes.txt
Perhaps you would also like to change your last script (start with cd /users/issues/slow_job_starts)
You will find more problems, like sed -e 's/^/${str}/' (the ${str} inside single quotes won't be replaced by a host), but this should get you started.
EDIT:
I added option -n to the ssh call.
Redirects stdin from /dev/null (actually, prevents reading from stdin).
Without this option only one node is checked.
I am trying to execute the following command in a Bash script:
grep 1001 -w /etc/passwd | cut -d ':' -f 1,4,5
grep 1004 -w /etc/passwd | cut -d ':' -f 1,4,5
it works fine from the command line in Linux, and if I remove the latter portion of the pipeline it executes properly from Bash as well.
here is my script thus far:
#/bin/bash
#find the group number correlated to reader and user
reader=`grep reader /etc/group | cut -d ":" -f3`
user=`grep user /etc/group | cut -d ":" -f3`
echo reader: $reader #prints 1004
echo user: $user #prints 1001
cmdRead="grep ${reader} -w /etc/passwd | cut -d \":\" -f 1,4,5"
cmdUser="grep ${user} -w /etc/passwd | cut -d \":\" -f 1,4,5"
echo executing command: ${cmdRead}
echo `${cmdRead}`
echo executing command: ${cmdUser}
echo `${cmdUser}`
the output of this code yields:
reader: 1004
user: 1001
executing command: grep 1004 -w /etc/passwd | cut -d ":" -f 1,4,5
grep: invalid argument ‘":"’ for ‘--directories’
Valid arguments are:
- ‘read’
- ‘recurse’
- ‘skip’
Usage: grep [OPTION]... PATTERN [FILE]...
Try 'grep --help' for more information.
executing command: grep 1001 -w /etc/passwd | cut -d ":" -f 1,4,5
grep: invalid argument ‘":"’ for ‘--directories’
Valid arguments are:
- ‘read’
- ‘recurse’
- ‘skip’
Usage: grep [OPTION]... PATTERN [FILE]...
Try 'grep --help' for more information.
I only started learning Bash yesterday so I apologize for the noob-esque question but greatly appreciate any help :)
Enclose your command in $( ... ), not quotes. Also, no need to quote the colon as the value of the -f parameter in cut, therefore no need to escape the quotes:
cmdRead=$(grep ${reader} -w /etc/passwd | cut -d: -f 1,4,5)
I would like to create a perl or bash script that will read keyboard input and assign a variable, perform a fixed string grep recursively within the current directory filled with Snort logs, and then automatically tcpdump the matched files, grep its output, and print the specified lines to the terminal. Does anyone have a good idea of how this should work?
Here is an example of the methodology I want from the script:
step 1: Read keyboard input and assign it to variable named string.
step 2 command: grep -Fr "$string"
step 2 output: snort.log.1470609906 matches
step 3 command: tcpdump -r snort.log.1470609906 | grep -F "$string" C-10
step 3 output:
Snort log
Here's some bash code that does that:
s="google.com"
grep -Frl "$s" | \
while IFS= read -r x; do
tcpdump -r "$x" | grep -F "$s" -C10
done
idk about perl but you can do it easily enough just in shell:
str="google.com"
find . -type f -name 'snort.log.*' -exec grep -FlZ "$str" {} + |
xargs -0 -I {} sh -c 'tcpdump -r "{}" | grep -F '"$str"' -C10'